Basic Review

tom.h.wilsontom.wilson@mail.wvu.edu

Department of Geology and GeographyWest Virginia University

Morgantown, WV

Let’s take a short tour of the Excel files provided by Waltham that are coordinated with text exercises and discussions

Click on this link Take a look at the following: Intro.xls

Exp.xls

Integ.xls

6’

8’

1. 2

2. 0

3. 5

4. 1

5. 5

6. 6

7. 2

8. 11

9. 3

10. 2

11. 2

Subscripts and superscripts provide information about specific variables and define mathematical operations.

k1 and k2 for example could be used to denote sedimentation constants for different areas or permeabilities of different rock specimens.

See Waltham for additional examples of subscript notation.

The geologist’s use of math often turns out to be a periodic and necessary endeavor. As time goes by you may find yourself scratching your head pondering once-mastered concepts that you suddenly find a need for.

This is often the fate of basic power rules.

Evaluate the following

xaxb =

xa / xb =

(xa)b =

xa+b

xa-b

xab

5

0

5 5log log5 | none of these

| | | ( ) log

( ) log | | | none of these

0 |1| | none of these

( ) | | | none of these

a

b c b c b c b

ff f ggg

b c b c b c bc

a a a

a a a a a ca b c a

a f g a a aa

a a

a a a a a

Question 1.2a Simplify and where possible evaluate the following expressions -

i) 32 x 34

ii) (42)2+2

iii) gi . gk

iv) D1.5. D2

Exponential notation is a useful way to represent really big numbers in a small space and also for making rapid computations involving large numbers - for example,

the mass of the earth is 5970000000000000000000000 kg

the mass of the moon is 73500000000000000000000 kg

Express the mass of the earth in terms of the lunar mass.

While you’re working through that with pencil and paper let me write down these

two masses in exponential notation.

ME = 597 x 1022kg

MM = 7.35 x 1022kg

The mass of the moon (MM) can also be written as 0.0735 x 1024kg

22

( ) 22

597 10 597

7.35 10 7.35E

E MM

M xM

M x

Hence, the mass of the earth expressed as an equivalent number of lunar masses is

=81.2 lunar masses

Write the following numbers in exponential notation (powers of 10)?

The mass of the earth’s crust is 28000000000000000000000kg The volume of the earth’s crust is 10000000000000000000 m3

The mass of the earth’s crust is 2.8 x 1022 kg The volume of the earth’s crust is 1 x 10 19 m3

=mass/volume = 2.8 x 103 kg/m3

Differences in the acceleration of gravity on the earth’s surface (and elsewhere) are often reported in milligals. 1 milligal =10-5 meters/second2.

This is basically a unit conversion problem - you are given a value in one system of units, and the relation of requested units (in this case milligals) to the given units (in this case meters/s2)

1 milligal = 10-5 m/s2 hence 1 m/s2 in terms of milligals is found by multiplying both sides of the above equation by

105

to yield 105 milligals= 1m/s2 - thusg=(9.8m/s2) x 105 milligals/(m/s2) =9.8 x 105 milligals

The earth gains mass every day due to collision with (mostly very small) meteoroids. Estimate the increase in the earth’s mass since its formation assuming that the rate of collision has been constant and that

yearsxA

daykgx

t

M

e9

5

105.4

106

M/ t is the rate of mass gainAe is the age of the earth (our t)

But, what is the age of the earth in days? days 1.6425x10

years10 5.4 365

12

9

xxyeardaysAE

1) What is the total mass gained?

2) Express the mass-gain as a fraction of the earth’s present day mass

kgxME241095.5

. . ( / )Ei e A M t

Total

Mass

Gained

AE is a

t

Fractional

Mass EE MxMx

x 724

17

1066.11095.5

10855.9

The North Atlantic Ocean is getting wider at the average rate vs of 4 x 10-2 m/y and has width w of approximately 5 x 106 meters.

1. Write an expression giving the age, A, of the North Atlantic in terms of vs and w.

2. Evaluate your expression to answer the question - When did the North Atlantic begin to form?

How thick was it originally?

Over what length of time was it deposited?

510,000 years

2.1 to 2.7 million years ago

0.5 to 2.1 million years ago

http://www.sciencedaily.com/releases/2008/04/080420114718.htm

http://www.nasa.gov/mission_pages/MRO/multimedia/phillips-20080515.html

The example presented on page 3 illustrates a simple age-depth relationship for unlithified sediments

depthxkAge This equation is a quantitative statement of what we all have an intuitive understanding of - increased depth of burial translates into increased age of sediments. But as Waltham suggests - this is an approximation of reality.

What does this equation assume about the burial process? Is it a good assumption?

Example - if k = 1500 years/m calculate sediment age at depths of 1m, 2m and 5.3m. Repeat for k =3000 years/m

1m

2m

5.3m

Age = 1500 years

Age = 3000 years

Age = 7950 years

For k = 3000years/m

Age = 3000 years

Age = 6000 years

Age = 15900 years

kzanotationsymbolic a=age, z=depthwhere

You probably recognized that the equation we started with

Depth x kAge

is the equation of a straight line.

bmxy The general equation of a

straight line is

In this equation -

which term is the slope and which is the intercept?

bmxy

Depth x kAgeIn this equation - which term is the slope and which is

the intercept?

line theof slope theis k

zero bemust intercept the

A more generalized representation of the age/depth relationship should include an intercept term -

0AD kA

The geologic significance of A0 - the intercept - could be associated with the age of the upper surface after a period of erosion. Hence the exposed surface of the sediment deposit would not be the result of recent

The slope of the line is, in this case, an inverse rate. Our dependant variable is depth, which would have units of meters or feet, for example. The equation defines depth of burial in terms of age. K, the slope transforms a depth into a number of years and k must have units of years/depth.

0

10000

20000

30000

40000

50000

AG

E (

year

s)

0 20 40 60 80 100

Depth (meters)

0AD kA

sedimentation but instead would be the remains of sediments deposited at an earlier time A0.

-50000

0

50000

100000

150000

AG

E (

year

s)

-20 0 20 40 60 80 100

depth (meters)

The slope of this line is t/x =1500years/meter, what is the intercept?

The intercept is the line’s point of intersection along the y (or Age) axis at depth =0.

t

x

0AD kA

If only the relative ages of the sediments are known, then for a given value of k (inverse deposition rate) we would have a family of possible lines defining age versus depth.

-10000

0

10000

20000

30000

40000

AG

E (

year

s) 50000

60000

70000

0 20 40 60 80 100

Depth (meters)

Are all these curves realistic?

What are the intercepts?

0AD kA

Consider the significance of A0 in the following context

If k is 1000 years/meter, what is the velocity that the lake bed moves up toward the surface?

If the lake is currently 15 meters deep, how long will it take to fill up?

Consider the case for sediments actively deposited in a lake.

0AD kA

What is the intercept?

The slope (k) does not change. We still assume that the thickness of the sediments continues to increase at the rate of 1 meter in 1000 years.

Hint: A must be zero when D is 15 meters

Age =?

-1x105

-5x104

0

5x104

1x105

Age

(ye

ars)

-50 0 50 100

Depth (meters)

You should be able to show that A0 is -15,000 years. That means it will take 15,000 years for the lake to fill up.

-15,000

present day depth at age = 0.

Our new equation looks like this -

15,000-D1000A

-1x105

-5x104

0

5x104

1x105

Age

(ye

ars)

-50 0 50 100

Depth (meters)

… we would guess that the increased weight of the overburden would squeeze water from the formation and actually cause grains to be packed together more closely. Thus meter thick intervals would not correspond to the same interval of time. Meter-thick intervals at greater depths would correspond to greater intervals of time.

0AD kA

Is this a good model?

We might also guess that at greater and greater depths the grains themselves would deform in response to the large weight of the overburden pushing down on each grain.

These compaction effects make the age-depth relationship non-linear. The same interval of depth D at large depths will include sediments deposited over a much longer period of time than will a shallower interval of the same thickness.

The relationship becomes non-linear.

The line y=mx+b really isn’t a very good approximation of this age depth relationship. To characterize it more accurately we have to introduce non-linearity into the formulation. So let’s start looking at some non-linear functions.

Quadratic vs. Linear Behavior

-50 0 50 100

Depth (meters)

-100000

-50000

0

50000

100000

150000

Age

Compare the functions

15,000-D1000Aand (in red)

15,000-D10003 2 DA

What kind of equation is this?

This is a quadratic equation. The general form of a quadratic equation is

cbxaxy 2

-6 -4 -2 0 2 4 6

X

-75

-25

25

75

125

Y

Quadratics

23 60y x

22xy

20102 2 xxy

One equation differs form that used in the text

The increase of temperature with depth beneath the earth’s surface is a non-linear process.

Waltham presents the following table

0

1000

2000

3000

4000

5000

T

0 1000 2000 3000 4000 5000 6000 7000

Depth (km)

We see that the variations of T with Depth are nearly linear in certain regions of the subsurface. In the upper 100 km the relationship

0

1000

2000

3000

4000

5000

T

0 1000 2000 3000 4000 5000 6000 7000

Depth (km)

Can we come up with an equation that will fit the variations of temperature with depth - for all depths?

Let’s try a quadratic.

1020 xT

101725.1 xTFrom 100-700km the relationship

provides a good approximation.

works well.

0

1000

2000

3000

4000

5000

T

0 1000 2000 3000 4000 5000 6000 7000

Depth (km)

The quadratic relationship plotted below is just one possible relationship that could be derived to

explain the temperature depth variations.

77.679528.1)10537.1( 24 xxxT

5000

3000

1000

T

0

2000

4000

0 1000 2000 3000 4000 5000 6000 7000

Depth (km)

0

1000

2000

3000

4000

5000

T

0 1000 2000 3000 4000 5000 6000 7000

Depth (km)

111005.1)10255.8( 25 xxxT68053.1)10537.1( 24 xxxT

The formula - below right - is presented by Waltham. In his estimate, he has not tried to replicate the variations of temperature in the

upper 100km of the earth.

5000

3000

1000

T

0

2000

4000

0 1000 2000 3000 4000 5000 6000 7000

Depth (km)

Either way, the quadratic approximations do a much better job than the linear ones, but, there is still significant error in the estimate of T for a given depth.

Can we do better?

0

1000

2000

3000

4000

5000

T(O

C)

0 1000 2000 3000 4000 5000 6000 7000

Depth (km)

To do so, we explore the general class of functions referred to as polynomials. A polynomial is an equation that includes x to the power 0, 1, 2, 3, etc. The straight line

cbxaxy 2

bmxy

is referred to as a first order polynomial. The order corresponds to the highest power of x present in the equation - in the above case the highest power is 1.

The quadratic is a second orderPolynomial, and the equation

021 ... acxbxaxy nnn

is an nth order polynomial.

021 ... acxbxaxy nnn

In general the order of the polynomial tells you that there are n-1 bends in the data or n-1 bends along the curve. The quadratic, for example is a second order polynomial and it has only one bend. But the number of bends in the data is not necessarily a good criteria for determining what order polynomial should be used to represent the data.

The temperature variations rise non-linearly toward a maximum value (there is one bend in the curve), however, the quadratic equation (second order polynomial) does not do an adequate job of defining these variations with depth.

Noting the number of bends in the curve might provide you with a good starting point. You could then increase the order to obtain further improvements.

5000

3000

1000

T

0

2000

4000

0 1000 2000 3000 4000 5000 6000 7000

Depth (km)

5000

3000

1000

T

0

2000

4000

0 1000 2000 3000 4000 5000 6000 7000

Depth (km)

93064.100031.01085.21012.1 238412 xxxxdxT

Waltham offers the following 4th order polynomial as a better estimate of temperature

variations with depth.

41.394054.300113.01099.110289.1 237411 xxxxdxT

0 1000 2000 3000 4000 5000 6000 7000

Depth (km)

0

1000

2000

3000

4000

5000

T

4th

Polynomial

Here is another 4th order polynomial which in this case attempts to fit the near-surface 100km. Notice that this 4th order equation (redline plotted in graph) has three bends or turns.

In sections 2.5 and 2.6 Waltham reviews negative and fractional powers. The graph below illustrates the set of curves that result as the exponent p in

0aaxy p

is varied from 2 to -2 in -0.25 steps, and a0 equals 0. Note that the negative powers rise quickly up along the y axis for values of x less than 1 and that y rises quickly with increasing x for p greater than 1.

X2

X1.75

X-2

X-1.75

x

0 1 2 3 4 50

200

400

600

800

1000

1200

Y

Power Laws

?01.0 isWhat

?01.0 isWhat

42

2

2

2

Power Laws - A power law relationship relevant to geology describes the variations of ocean floor depth as a function of distance from a spreading ridge (x).

02/1 daxd

Spreading Ridge

0 200 400 600 800 1000

X (km)

0

1

2

3

4

5

D (km)

Ocean Floor Depth

What physical process do you think might be responsible for this pattern of seafloor subsidence away from the spreading ridges?

Section 2.7 Allometric Growth and Exponential Functions

Allometric - differential rates of growth of two measurable quantities or attributes, such as Y and X, related through the equation Y=abX -

This topic brings us back to the age/thickness relationship. Earlier we assumed that the length of time represented by a certain thickness of a rock unit, say 1 meter, was a constant for all depths. However, intuitively we argued that as a layer of sediment is buried it will be compacted - water will be squeezed out and the grains themselves may be deformed. The open space or porosity will decrease.

Waltham presents us with the following data table -

Over the range of depth 0-4 km, the porosity decreases from 60% to 3.75%!

Depth

0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00

Por

osit

y

0.0

0.1

0.2

0.3

0.4

0.5

0.6

This relationship is not linear. A straight line does a poor job of passing through the data points. The slope (gradient or rate of change) decreases with increased depth.

zx 2 6.0

Waltham generates this data using the following relationship.

zx 2 6.0This equation assumes that the initial porosity (0.6) decreases by 1/2 from one kilometer of depth to the next. Thus the porosity () at 1 kilometer is 2-1 or 1/2 that at the surface (i.e. 0.3), (2)=1/2 of (1)=0.15 (i.e. =0.6 x 2-2 or 1/4th of the initial porosity of 0.6.

Equations of the type

cxaby

are referred to as allometric growth laws or exponential functions.

(0.6)2 z In the equation

a = ?

b = ?

c = ?

cxaby

The constant b is referred to as the base.

Recall that in relationships like y=10a, a is the power to which the base 10 is raised in order to get y.

a=0.6

b=2

c= -1

The porosity-depth relationship is often stated using a base different than 2. The base which is most often used is the natural base e and e equals 2.71828 ..

In the geologic literature you will often see the porosity depth relationship written as

-cz0 e

0 is the initial porosity, c is a compaction factor and z - the depth.

Sometimes you will see such exponential functions written as -cz

0 exp

In both cases, e=exp=2.71828

z

-

0 e

Waltham writes the porosity-depth relationship as

Note that since z has units of kilometers (km) that c must have units of km-1 and must have units of km.

z

-

0 eNote that in the above form

when z=,

01-

0

-

0 368.0

ee

represents the depth at which the porosity drops to 1/e or 0.368 of its initial value.

-cz0 e In the form c is the reciprocal of that depth.

Logarithms

Above, when we talked about functions like cxaby and

cxay 10b and 10 are what we refer to as bases. These are constants and we can define any other number in terms of these constants raised to a certain power.

xyei 10 .. Given any number y, we can express y as 10 raised to some power x

Thus, given y =100, we know that x must be equal to 2.

xy 10By definition, we also say that x is the log of y, and can write

xy x 10loglog

So the powers of the base are logs. “log” can be thought of as an operator like x and which yields a certain result. Unless otherwise noted, the operator “log” is assumed to represent log base 10. So when asked what is

45y where,log yWe assume that we are asking for x such that

4510 x

Sometimes you will see specific reference to the base and the question is written as

45y where,log10 y

y10log leaves no room for doubt that we are specifically interested in the log for a base of 10.

ypow 10

One of the confusing things about logarithms is the word itself. What does it mean? You might read log10 y to say -”What is the power that 10 must be raised to to get y?” How about this operator? -

The power of base 10 that yields () y

653.1log10 y

What do you think? Are small earthquakes much more common than large ones?

Fortunately, the answer to this question is yes, but is there a relationship between the size of an earthquake and the number of such earthquakes?

5 6 7 8 9 10

Richter Magnitude

0

100

200

300

400

500

600

Num

ber

of e

arth

quak

es p

er y

ear

Observational data for earthquake magnitude (m) and frequency (N, number of earthquakes per year with magnitude greater than m)

What would this plot look like if we plotted the log of N versus m?

0.01

0.1

1

10

100

1000

Num

ber

of e

arth

quak

es p

er y

ear

5 6 7 8 9 10

Richter Magnitude

This looks like a linear relationship. Recall the formula for a straight line?

bmxy

0.01

0.1

1

10

100

1000

Num

ber

of e

arth

quak

es p

er y

ear

5 6 7 8 9 10

Richter Magnitude

What would y be in this case?

Ny log

What is b?

the intercept

5 6 7 8 9 10

Richter Magnitude

0.01

0.1

1

10

100

1000

Num

ber

of e

arth

quak

es p

er y

ear

cbmN log

The Gutenberg-Richter Relation

-b is the slope and c is the intercept.

0 50 100 150 200 250 300 350 400

Square Root of Fault Plane Area (kilometers) (Characteristic Linear Dimension)

-2

-1

0

1

2

3

Log

of

the

Num

ber

of E

arth

quak

es p

er Y

ear

-2

-1

0

1

2

3

Log

of

the

Num

ber

of E

arth

quak

es p

er Y

ear

1 10 100 1000

Square Root of Fault Plane Area (kilometers) (Characteristic Linear Dimension)

In passing, also note that the magnitude axis can be replaced with the square root of fault plane area since earthquake magnitude is proportional to the square root of area of the fault plane along which rupture occurred. This relationship is linear in the log-log plot shown above right.

Based on the preceding comment, it should be no surprise that we also find a similar relationship between the number of faults of length greater than or equal to a given size at a particular outcrop location

0

2000

4000

6000

8000

10000

12000

N

0 2 4 6 8 10

Fault Length (meters)

Another nice attribute of logs is that when you plot the log of y (i.e. the power that 10 (or some other base) has to be raised to to yield y) rather than y itself, it is easier to see relative differences in the value y.

For example, replotting the N vs L data on logN vs. log(L) scale we get a straight line and can easily see the relationship of differences in N relative to size L.

1

10

100

1000

10000

N

0.001 0.01 0.1 1 10

Fault Length (meters)

In this case, we see that the log-log relationship is linear.

10-1-2-3exponent only

1

10

100

1000

10000

N

10-3

10-2

10-1

100

101

Fault Length (meters)

Sometimes the labels on the axes will consist only of the exponent (log or power) itself. Thus, as shown below, the gridlines are labeled 100 or just 0, 101 or just 1, etc.

5 6 7 8 9 10

Richter Magnitude

0.01

0.1

1

10

100

1000

Num

ber

of e

arth

quak

es p

er y

ear cbmN log

The Gutenberg-Richter Relation

is linear in a log-log format because m is earthquake magnitude and you have heard that an earthquake magnitude of 5, for example, represents ground motion whose amplitude is 10 times that associated with a magnitude 4 earthquake.

One of the most commonly used “Richter magnitude” scales determines the magnitude of shallow earthquakes from surface waves according to the following equation

3.3log66.1log10 T

Am

where T is the period in seconds, A the maximum amplitude of ground motion in m (10-6 meters) and is the epicentral distance in degrees between the earthquake and the observation point.

We’ve already worked with three bases - 2, 10 and e. Whatever the base, the logging operation is the same.

10.get to toraised bemust 5 power that theis what asks 10log5

? 10log5

How do we find these powers?

5log

10log 10log

10

105

431.1699.0

1 10log5

105 431.1 thus

In general, base

numberbase

10

10

log

)(log number) some(log

or

b

ab

10

10

log

)(log alog

Try the following on your own

?3log

)7(log 7log

10

103

8log8

21log7

7log4

subscript no with log, asten often writ is log that find You will 10

log10 is referred to as the common logarithm

ln. asten often writ is log e

2.079 ln8 8log e

thus

loge or ln is referred to as the natural logarithm. All other bases are usually specified by a subscript on the log, e.g.

etc. ,logor og 25l

Finish reading Chapters 1 and 2 (pages 1 through 38) of Waltham

After we finish our basic review, we will learn how to use Excel a scientific computing and graphing software package to solve some problems related to the material covered in Chapters 1 and 2.